Peano Theorem: Let $f$ be continuous in $\Omega=I_a\times B_b$, where $I_a=\{t:|t-t_0|\leq a\}$ and $B_b=\{x:|x-x_0|\leq b\}$. If $|f|<M$ in $\Omega$, the $$x'=f(t,x) \qquad x(t_0)=x_0$$ has at least one solution in $I_\alpha$, where $\alpha=\min{\{a,b/M\}}$.
Changing the condition $|f|<M$ by $|f|\leq M$, is it possible to obtain the same results as the theorem above?
Edit: I thought I had managed to solve the problem in the post below. But as I commented on the post itself, I think I found an error, can someone help me?
With the help of user Anonymous, it was possible to solve the exercise and, with the change of the hypothesis, arrive at the same conclusion as the stated theorem.
Assuming: $a\geq b/M$
By hypothesis, we have $|f|\leq M$, thus $|f|<2M$. In this way, the Cauchy problem $$x'=f(t,x) \qquad x(t_0)=x_0$$ satisfies the hypotheses of Peano theorem. That is, there is a solution $x(t)$ in the interval $[t_0 -b/2M,t_0+b/2M]$.
Likewise, the following Cauchy problems have a solution. (See comment below)
$$y_+'=f(t,y_+)\qquad y_+(t_1)=x(t_1) \quad\text{where } t_1=t_0+b/2M$$ $$y_-'=f(t,y_-)\qquad y_-(t_2)=x(t_2) \quad\text{where } t_2=t_0-b/2M$$
Therefore $y_+$ is defined in $[t_0,t_0+b/M$] and $y_-$ in $[t_0 - b/M, t_0]$.
$$\varphi(t) = \left\{ \begin{array}{ll} y_-(t) & \mbox{if } t\in [t_0 - b/M, t_0 -b/2M) \\ x(t) & \mbox{if } t\in [t_0 -b/2M,t_0+b/2M] \\ y_+(t) & \mbox{if } t\in (t_0+b/2M, t_0 +b/M) \\ \end{array} \right.$$
Thus, $\varphi$ is a solution of $x'=f(t,x), x(t_0)=x_0$ defined in the interval $I_\alpha$ where $\alpha=\min\{a,b/M\}=b/M$
Assuming: $a> b/M$
Let $\epsilon=b/a-M>0$, thus $a=b/(M+\epsilon)$. Then $\vert f\vert<M+\epsilon$, so by the Peano theorem, there is a solution $x(t)$ over $I_\alpha$ where $\alpha=\min\{a,b/(M+\epsilon)\}=a$. But we also know $a=\min\{a,b/M\}$ since $a<b/M$, so in fact $\alpha=\min\{a,b/M\}$.
Comment: I believe that I may have made a mistake in assuming that the solution exists. In Peano's theorem, $t_0$ and $x_0$ are required to be in the domain of $f$. I was able to guarantee that $t_1$ is in $I_a$. Because taking $a'=a-b/2M$ it is concluded that $I_{a'}\subset I_{a}$ and $\min\{a',b/2M\}=b/2M$.
But the problem is ensuring that $y_+(t_1)=x(t_1) \in B_b$. Is it possible to correct this proof to obtain the conclusion of the theorem?